Organic I (Reactions)

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1 Alkanes Free-radical substitution Halogenation of alkanes is a free radical reaction. It results in H bonds being replaced for Hal bonds. The reagents and conditions are to mix the alkane with bromine or chlorine and expose to ultra-violet light. The ultra-violet light provides enough energy to break the halogen bond, which forms species called free radicals. When the Hal-Hal bond breaks, it does so with one electron from the bond going with each atom. This is what leads to production of the free-radicals mentioned. Definition: Homolytic fission is when a bond breaks and each atom in the bond takes one electron from the bond. Definition: A free-radical is a species with an unpaired electron. Definition: A substitution reaction is one in which an atom or group of atoms in a molecule is replaced by another atom or group of atoms. Free radicals are very reactive species, as energy is released when they pair up their electrons to make new covalent bonds. CH 4 + Cl 2 CH 3 CH 3 + Cl 2 CH 3 Cl + HCl CH 3 CH 2 Cl + HCl Free radical reactions such as this one are often very random and unpredictable. For example, there is no way on controlling which H bonds are replaced, and further substitution may also occur CH 3 Cl + Cl 2 CH 2 Cl 2 + HCl So, a mixture of products is obtained, which makes this an inefficient synthesis of halogenoalkanes. In addition, mixtures of alkanes and chlorine are often explosive when radiated with ultra-violet light. Combustion Complete combustion of alkanes gives CO 2 and H 2 O. CH 4 + 2O 2 CO 2 + 2H 2 O These reactions are very exothermic and burning alkanes and other hydrocarbons are commonly used as fuels. Page 1 of 10

3 Alkenes Addition reactions Alkenes undergo addition reactions with hydrogen, halogens (such as bromine), hydrogen halides and potassium manganate(vii). Definition: An addition reaction is one in which two species combine to make single product. The reactions with bromine and potassium manganate are used as tests for alkenes. Tests for alkenes Test 1 Mix the alkene with a solution of bromine. Decolourisation, from brown to colourless. Bromine can be used either in organic solvents such as tetrachloromethane or hexane, alternatively it may be used in aqueous solution. In any case, the positive result is decolorisation of the solution. It would be good to indicate some practical awareness of the mixing would occur: gaseous alkenes bubble alkene through bromine solution. liquid alkene shake alkene with bromine solution. solid alkene, dissolve alkene in inert solvent, then shake with bromine solution. If bromine water is used, then the product obtained is a bromo-alcohol, as water participates in the reaction as a nucleophile in the second step of the mechanism. CH 3 CH=CH 2 + Br 2 + H 2 O CH 3 CH(OH)CH 2 Br + HBr If bromine in an inert solvent is used, then the product obtained is a dibromoalkane. CH 3 CH=CH 2 + Br 2 CH 3 CHBrCH 2 Br These addition reactions could also be classified as an oxidation of the alkene. Test 2 Mix the alkene a cold dilute aqueous solution of alkaline potassium manganate (VII). Decolourisation, from purple to green to colourless. A brown precipitate of MnO 2 forms. In this test, the alkene is turned into a diol. CH 3 CH=CH 2 + H 2 O + [O] CH 3 CH(OH)CH 2 OH KMnO 4 reacts with many substances, primary and secondary alcohols, aldehydes and even some ketones. What makes this test specific to alkenes is the conditions. When the KMnO 4 is alkaline, and cold, and dilute, it is such a poor oxidising agent that only alkenes are sufficiently reducing to react with it. This addition reaction can also be classified as an oxidation of the alkene. Page 3 of 10

4 Other reactions of alkenes: Hydrogen adds to alkenes to form alkanes. Reagents and conditions are pass the gases over a heated nickel catalyst. CH 3 CH=CH 2 + H 2 CH 3 CH 2 CH 3 This can also be classified as a reduction of the alkene. Hydrogen halides add to alkenes to form haloalkanes. Reagents and conditions are mix the gases at room temperature. CH 3 CH=CH 2 + HBr CH 3 CHBrCH 3 When the substance attacking the double bond is unsymmetrical (like HBr), the hydrogen atom goes on the atom with more H atoms to begin with. This will be studied more in a later unit, just remember it for now. It is Markovnikov s rule. Substances that add to C=C bonds, such as H 2, Br 2, and HBr are termed electrophiles. Definition: An electrophile is an electron deficient species that can accept a pair of electrons from an electron rich region of another molecule. Naming of alkenes The suffix ene is used to indicate that a compound contains a C=C bond, sometimes with a number to indicate where it is in the carbon chain. The direction of numbering along the chain is chosen so as to give smaller numbers to the carbons containing the double bonds. CH 2 =CH 2 CH 3 CH=CH 2 CH 3 CH 2 CH=CH 2 CH 3 CH=CHCH 3 ethene propene but-1-ene but-2-ene Draw ethene, showing all the bonds. If there are alkyl groups or halogen atoms to denote in the name, then the position of the double bond takes precedence. CH 2 =CHCH 2 CH 2 CH 2 C(CH 3 ) 3 6,6-dimethyl hept-1-ene Multiple double bonds are named as follows: CH 2 =CHCH=CH 2 CH 2 =C=CHCH 2 buta-1,3-diene buta-1,2-diene This a in buta is just to make the name easier to say, and you can probably forget about it. Page 4 of 10

5 Halogeno alkanes Substitution reactions Definition: A substitution reaction is one in which an atom or group of atoms in a molecule is replaced by another atom or group of atoms. Definition: A nucleophile is an electron rich species that is attracted to an electron deficient region of a molecule and can donate a pair of electrons to form a new covalent bond. The important point is the ability to donate a pair of electrons. Substitution with hydroxide (hydrolysis) The reagents and conditions for hydrolysis of halogenoalkanes are boil under reflux with dilute aqueous sodium hydroxide. CH 3 CH 2 Br + NaOH CH 3 CH 2 OH + NaBr Substitution with cyanide (cyanolysis) This just like hydrolysis but the nucleophile is CN -. The reagents and conditions are to boil the halogenoalkane under reflux with sodium or potassium cyanide in a solvent such as ethanol and water, or propanone. CH 3 CH 2 Br + NaCN CH 3 CH 2 CN + NaBr Substitution with ammonia (aminolysis) This is just like hydrolysis but with ammonia as the nucleophile. The reagents and conditions are heat the halogenoalkane with excess concentrated ammonia in ethanol in a sealed tube. CH 3 CH 2 Br + NH 3 CH 3 CH 2 Br + 2NH 3 CH 3 CH 2 NH 2 + HBr, or CH 3 CH 2 NH 2 + NH 4 Br In the above reactions, the OH -, CN - and NH 3 molecules are behaving as nucleophiles. Page 5 of 10

6 Halogeno alkanes Elimination reactions Definition: An elimination reaction is one where the elements of a small molecule are removed from a larger one. Halogenoalkanes undergo elimination by removal of the halogen atom and a hydrogen atom. The hydrogen atom must come from a carbon atom that is adjacent to the carbon atom that loses the halogen atom. The C=C is formed between the two atoms that lose the hydrogen and halogen atoms. Often there is more than one hydrogen atom that can be removed, in which case the product will be a mixture of molecules. Sometimes the products can exist as cis and trans isomers. The reagents and conditions for elimination are heat strongly under reflux with concentrated potassium or sodium hydroxide in ethanol. Under these conditions, the OH - ion behaves as a base by removing a hydrogen atom from the molecule. CH 3 CH 2 Br + NaOH CH 2 =CH 2 + H 2 O + NaBr In the above equation, the elements of a small molecule that have been removed are those of HBr, although an actual molecule of HBr is not created. When more than one product is possible the one with the greater number of alkyl groups present on the C=C is formed in greater yield. When geometrical isomers are possible, cis and trans are formed in equal yield. CH 3 CH 2 CHBrCH 3 CH 3 CH 2 CH=CH 2 (smaller yield) CH 3 CH = CHCH 3 (greater yield, cis and trans 1:1) Halogenoalkanes are more prone to elimination if they are more branched, i.e. tertiary eliminates more easily than secondary than primary. Branching provides more hydrogen atoms for the reaction to start with, and also sterically hinders the carbon atom thus preventing the OH - from acting as a nucleophile and causing a substitution reaction instead. Very important point: In substitution the OH - behaves as a nucleophile (electron pair donor), whereas in elimination it behaves as a base (proton acceptor). Draw 2-bromobutane showing all bonds. Label the H atoms that can be lost to form but- 1-ene and those that can be lost to form but-2-ene. Page 6 of 10

7 Test for halogen atoms in halogeno alkanes, Cl, Br and I Step one - Free the halide ion from the organic molecule. How this is done depends on the type of molecule, but for simple halogeno alkanes it is done by heating the halogenoalkane under reflux with dilute, aqueous sodium hydroxide solution. Step two add excess dilute nitric acid, to neutralise excess NaOH or other alkaline impurities. Any alkaline impurities will give a grey precipitate when silver nitrate is added, which could be interpreted as a false-positive result. Step three add silver(i)nitrate solution. Chloride ions will give a white precipitate Bromide ions will give a cream precipitate Iodide ions will give a yellow precipitate Step four filter off the precipitate and test its solubility in ammonia solution: 1. AgCl will dissolve in dilute ammonia solution 2. AgBr will dissolve in concentrated ammonia solution 3. AgI will not dissolve in any concentration of ammonia solution. See group VII notes for more details of this test for halide ions. Naming of halogenoalkanes Halogen atoms are specified with a prefix and a sometimes a number to specify where the halogen is on the carbon chain. The chain is numbered from the direction that gives the smaller number of set of numbers for atoms which have halogen atoms attached. Halogen F Cl Br I CH 3 Cl CH 3 CHBrCH 3 CH 3 CHBrCH 2 CH 2 CH 3 Prefix Fluoro Chloro Bromo Iodo chloromethane 2-bromopropane 2-bromopentane The prefixes are modified if more than one halogen is present. They are given alphabetically according the name of the halogen. Alkyl groups can also be included. CH 3 CBr 2 CBrCl 2 CH 3 CH(CH 3 )CHF 2 1,2,2-tribromo-1,1-dichloro propane 1,1-difluoro-2-methyl propane Page 7 of 10

8 Alcohols Oxidation of alcohols Reagents and conditions are to heat with potassium dichromate solution acidified with dilute aqueous sulphuric acid. You must specify the acid, writing H + will not get you full marks. Primary alcohols are oxidised to aldehydes then carboxylic acids. CH 3 CH 2 OH CH 3 CHO CH 3 CO 2 H In order to obtain the aldehyde, the reaction is carried out with excess alcohol and the aldehyde is distilled off as it is formed. This is possible because the aldehyde has the lowest boiling point of the three compounds, because it does not have hydrogen bonds between its molecules. Reagents and conditions are add potassium dichromate(vi) in dilute sulphuric acid to the hot alcohol and allow the aldehyde to distil off. In order to obtain the carboxylic acid, the reaction is carried out with excess oxidising agent and the heating is carried out under reflux. This is in order to maximise yield. Reagents and conditions are boil the alcohol with excess potassium dichromate(vi) in dilute sulphuric acid under reflux. In organic oxidation reactions, equations can be balanced using the symbol [O]. CH 3 CH 2 OH + [O] CH 3 CHO + [O] CH 3 CHO + H 2 O CH 3 CO 2 H Secondary alcohols are oxidised to ketones. CH 3 CH(OH)CH 3 + [O] CH 3 COCH 3 + H 2 O Tertiary alcohols are not oxidised under these conditions. Reaction of alcohols with dehydrating agents Alcohols can be dehydrated to form alkenes. The reactions can be classified as dehydration or elimination. Reagents and conditions are heat under reflux at 170 o C with excess conc. H 2 SO 4, or heat at 300 o C over Al 2 O 3. You should be able to predict the alkenes that would form from a typical alcohol, such as butan-2-ol. The OH group is removed, along with any hydrogen atom attached to a carbon next to the carbon with the OH group. Note that geometrical isomers may form. CH 3 CH 2 CH(OH)CH 3 Page 8 of 10 CH 3 CH=CHCH 3, major product, cis and trans. CH 3 CH 2 CH=CH 2, minor product.

9 Reaction of alcohols with halogenating agents Halogenating agents replace the OH group with a halogen. You need to be familiar with three different ones for this unit. Method one PCl 5 (s) Reagents and conditions are carefully add solid PCl 5 at room temperature, under dry conditions. CH 3 CH 2 OH + PCl 5 CH 3 CH 2 Cl + POCl 3 + HCl Method two NaBr(s) and conc. H 2 SO 4 NaBr(s) and conc. H 2 SO 4 react to form HBr (but see group VII notes). Reagents and conditions add a mixture of the alcohol and conc. H 2 SO 4 to NaBr at room temperature. CH 3 CH 2 OH + HBr CH 3 CH 2 Br + H 2 O Method three phosphorous and iodine Phosphorous and iodine react together to form PI 3. Reagents and conditions are add the alcohol to moist red phosphorous and iodine at room temperature. 3CH 3 CH 2 OH + PI 3 3CH 3 CH 2 I + H 3 PO 3 Test for OH groups in alcohols (and carboxylic acids). The PCl 5 reaction above is used to test for an OH group. Note that it works for any compound that contains an OH group, so both alcohols and carboxylic acids, so it is not a test for an alcohol specifically. Water also contains an OH group, so the conditions must be anhydrous to avoid a false-positive result. Carefully add a small amount of solid PCl 5 to a small amount of the sample in a clean dry test tube, in a fume cupboard at room temperature. The sample must be dry. Effervescence of steamy fumes (HCl) which turn moist blue litmus paper red and give a dense white smoke (NH 4 Cl) with an ammonia stopper. PCl 5 + ROH POCl 3 + RCl + HCl(g) HCl(g) + NH 3 (g) NH 4 Cl(s) Using dichromate as a test for alcohols. If heating a test substance with excess potassium dichromate(vi) in dilute sulphuric acid turns it green, then the test substance would be a primary or secondary alcohol, but remember, it could also be an aldehyde. Page 9 of 10

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